is it possible for a rectangle to have its adjacent sides of length 2cm and 4cm and diagonals bisecting each other st right angles give reasons?
Answers
Step-by-step explanation:
Answer:
\bf \underline{The\ original\ number\ is\ 92.}
The original number is 92.
Step-by-step explanation:
Let the digit at tens and units place be x and y respectively.
Amd the number be (10x+y)
Given
Product of digit is 18.
So,
ATE, xy=18
or y=18/x -(1)
Also, when 63 is subtracted from the number, the digits interchange their place.
So, the number formed by interchanging the digits is (10y+x)
ATE,
(10x+y)-63=(10y+x)
10x-x+y-10y=63
9x-9y=63
9(x-y)=63
x-y=7 -(2)
Using y=18/x in eq. (2)
\begin{lgathered}x-\frac{18}{x}=7\\ \\ \implies \rm x^2-18=7\\ \\ \implies \rm \frac{x^2-18}{x}=7\\ \\ \implies \rm x^2-7x-18=0\\ \\ \implies \rm x^2-9x+2x-18=0\\ \\ \implies \rm x(x-9) 2(x-9)=0\\ \\ \implies \rm (x+2) (x-9)=0\\ \\ \implies \bf x=-2\ or\ \bf 9 \\ \rm But\ digit\ can't\ be\ negative.\\ \therefore \bf x=9\end{lgathered}
x−
x
18
=7
⟹x
2
−18=7
⟹
x
x
2
−18
=7
⟹x
2
−7x−18=0
⟹x
2
−9x+2x−18=0
⟹x(x−9)2(x−9)=0
⟹(x+2)(x−9)=0
⟹x=−2 or 9
But digit can
′
t be negative.
∴x=9
Putting x=9 in eq. (1)
9y=18
y=2
So, the number is (10×9+2)=90+2=92
Answer:
no can't possible to have a rectangle it's adject sides are 2 cm and 4 cm